We are asked to simplify the expression $\frac{x^2-2x-15}{x^2-9} \times \frac{x^2-8x+15}{x^2-10x+25}$.

1. Problem Description

We are asked to simplify the expression $\frac{x^2-2x-15}{x^2-9} \times \frac{x^2-8x+15}{x^2-10x+25}$.

2. Solution Steps

First, we factor each of the quadratic expressions.

$x^2 - 2x - 15 = (x-5)(x+3)$

$x^2 - 9 = (x-3)(x+3)$

$x^2 - 8x + 15 = (x-5)(x-3)$

$x^2 - 10x + 25 = (x-5)(x-5) = (x-5)^2$

So the original expression becomes

$\frac{(x-5)(x+3)}{(x-3)(x+3)} \times \frac{(x-5)(x-3)}{(x-5)(x-5)}$

Now, we can simplify by canceling common factors. We cancel $(x+3)$, $(x-3)$, and $(x-5)$.

$\frac{(x-5)(x+3)}{(x-3)(x+3)} \times \frac{(x-5)(x-3)}{(x-5)(x-5)} = \frac{(x-5)}{(x-5)} = 1$

However, $x$ cannot be $5$, $3$ or $-3$ because this would make the denominator zero. Therefore, the simplified expression is 1, given the conditions that $x \neq 5$, $x \neq 3$, and $x \neq -3$.

$\frac{(x-5)(x+3)}{(x-3)(x+3)} \times \frac{(x-5)(x-3)}{(x-5)^2} = \frac{(x-5)(x+3)(x-5)(x-3)}{(x-3)(x+3)(x-5)(x-5)} = \frac{(x-5)^2(x+3)(x-3)}{(x-3)(x+3)(x-5)^2} = 1$

3. Final Answer

1